1. For motion up the incline

The block is projected upward on a rough incline of angle $\theta = 30^\circ$ with initial speed $v_0$.

While moving up, both gravity component and friction act down the plane.

So retardation is $a_1 = g\sin\theta + \mu g\cos\theta$

Using $v^2 = u^2 - 2a_1 s$ when it comes to rest at the highest point, $0 = v_0^2 - 2(g\sin\theta + \mu g\cos\theta)s$

Hence, $s = \frac{v_0^2}{2g(\sin\theta + \mu\cos\theta)} \qquad (1)$

2. For motion down the incline

From the highest point it slides back down a distance $s$ and reaches the starting point with speed $\dfrac{v_0}{2}$.

Now gravity component acts down the plane, friction acts up the plane.

So acceleration down the plane is $a_2 = g\sin\theta - \mu g\cos\theta$

Using $v^2 = 2a_2 s$ (since it starts from rest at the top), $\left(\frac{v_0}{2}\right)^2 = 2(g\sin\theta - \mu g\cos\theta)s$

Substitute $s$ from (1): $\frac{v_0^2}{4} = 2(g\sin\theta - \mu g\cos\theta)\cdot \frac{v_0^2}{2g(\sin\theta + \mu\cos\theta)}$

Simplify: $\frac{1}{4} = \frac{\sin\theta - \mu\cos\theta}{\sin\theta + \mu\cos\theta}$

3. Put $\theta = 30^\circ$

We know $\sin 30^\circ = \frac12, \qquad \cos 30^\circ = \frac{\sqrt3}{2}$

So $\frac{1}{4} = \frac{\frac12 - \mu\frac{\sqrt3}{2}}{\frac12 + \mu\frac{\sqrt3}{2}}$

Multiply numerator and denominator by $2$: $\frac{1}{4} = \frac{1 - \mu\sqrt3}{1 + \mu\sqrt3}$

Cross-multiply: $1 + \mu\sqrt3 = 4(1 - \mu\sqrt3)$ $1 + \mu\sqrt3 = 4 - 4\mu\sqrt3$ $5\mu\sqrt3 = 3$ $\mu = \frac{3}{5\sqrt3} = \frac{\sqrt3}{5}$

Numerically, $\mu \approx \frac{1.732}{5} = 0.3464$

Given $\mu \approx \frac{I}{1000}$ so $I \approx 346.4$

Nearest integer: $\boxed{346}$

4. Comparison with stored answer

Stored correct answer = $346$.

Our derived answer is also $346$, so it agrees.