1. Given data

• Mass of block: $m = \sqrt{3}\,\text{kg}$
• Coefficient of friction: $\mu = \frac{1}{3\sqrt{3}}$
• Force is applied on the vertical face at $60^\circ$ with the horizontal.
• $g = 10\,\text{m/s}^2$

We need the critical force so that the block is just about to move, and it is given as $3x$.

2. Resolve the applied force

Let the applied force be $F$.

Since it is applied at $60^\circ$ with the horizontal:

• Horizontal component: $F_x = F\cos 60^\circ = \frac{F}{2}$
• Vertical component: $F_y = F\sin 60^\circ = \frac{\sqrt{3}}{2}F$

Because the force is applied on the vertical surface, the horizontal component presses the block against the rough horizontal surface, while the vertical component tends to lift the block upward.

So the normal reaction is reduced by the upward component.

3. Normal reaction

Weight of the block: $mg = \sqrt{3}\times 10 = 10\sqrt{3}\,\text{N}$

Vertical equilibrium at the limiting condition:

$N = mg - F\sin 60^\circ$

So, $N = 10\sqrt{3} - \frac{\sqrt{3}}{2}F$

4. Limiting friction

Maximum static friction: $f_{\max} = \mu N$

Thus, $f_{\max} = \frac{1}{3\sqrt{3}}\left(10\sqrt{3} - \frac{\sqrt{3}}{2}F\right)$

Simplify: $f_{\max} = \frac{1}{3}\left(10 - \frac{F}{2}\right)$

$f_{\max} = \frac{10}{3} - \frac{F}{6}$

5. Condition for critical force

For the block to remain at rest, the upward component $F\sin 60^\circ$ is balanced by weight and friction.

At the critical value for upward impending motion, friction acts downward at its maximum value.

So, $F\sin 60^\circ = mg + f_{\max}$

Substitute: $\frac{\sqrt{3}}{2}F = 10\sqrt{3} + \left(\frac{10}{3} - \frac{F}{6}\right)$

But this mixes irrational and rational terms incorrectly because friction is vertical force in newtons and must be computed directly without cancelling prematurely in that form. So let us write friction more carefully:

$f_{\max} = \mu N = \frac{1}{3\sqrt{3}}\left(10\sqrt{3} - \frac{\sqrt{3}}{2}F\right) = \frac{10}{3} - \frac{F}{6}$

Now vertical balance at limiting upward motion is: $\frac{\sqrt{3}}{2}F = 10\sqrt{3} + \left(\frac{10}{3} - \frac{F}{6}\right)$

This clearly cannot be right dimensionally with mixed surd terms from weight and friction, which indicates the interpretation is wrong.

6. Correct interpretation of the figure/setup

The force is applied on the vertical face of the block, so the horizontal component pushes the block along the rough horizontal plane and the vertical component changes the normal.

Thus:

• Horizontal component $F\cos 60^\circ = F/2$ tends to move the block horizontally.
• Vertical component $F\sin 60^\circ$ acts downward or upward depending on the figure. For the standard configuration here, it acts downward on the block, increasing normal reaction.

Then, $N = mg + F\sin 60^\circ = 10\sqrt{3} + \frac{\sqrt{3}}{2}F$

Maximum friction: $f_{\max} = \mu N = \frac{1}{3\sqrt{3}}\left(10\sqrt{3} + \frac{\sqrt{3}}{2}F\right)$

$f_{\max} = \frac{1}{3}\left(10 + \frac{F}{2}\right) = \frac{10}{3} + \frac{F}{6}$

For limiting horizontal motion, $F\cos 60^\circ = f_{\max}$

$\frac{F}{2} = \frac{10}{3} + \frac{F}{6}$

Multiply by $6$: $3F = 20 + F$

$2F = 20$

$F = 10\,\text{N}$

7. Find $x$

Given critical force $= 3x$, $3x = 10$

$x = \frac{10}{3} = 3.33$

8. Final answer

$\boxed{x = \frac{10}{3} \approx 3.33}$