1. Set up the geometry and forces

   A uniform cube of side $a$ rests on a rough horizontal floor.

   • Weight: $mg$, acting through the centre of mass.
   • A horizontal force $F$ is applied at a point whose height above the centre of mass is $b$.
   • Coefficient of friction with floor: $\mu = 0.4$.

   Since the centre of mass is at height $a/2$ above the floor, the force is applied at height $h = \frac{a}{2} + b$ above the floor.

2. Condition for the box to start sliding

   The minimum force needed to make the box move by sliding is when static friction reaches its limiting value: $F_{\text{slide}} = \mu mg$

3. Condition for the box to topple

   The box will begin to topple about the lower edge on the side of pulling when the torque of $F$ about that edge equals the restoring torque of weight.

   Taking moments about the bottom edge on the tipping side:

   • Moment arm of $F$: $h = \dfrac{a}{2}+b$
   • Moment arm of $mg$: $\dfrac{a}{2}$

   So, at toppling threshold, $F_{\text{topple}}\left(\frac{a}{2}+b\right)=mg\left(\frac{a}{2}\right)$

   Hence, $F_{\text{topple}}=\frac{mg\,(a/2)}{(a/2+b)}$

4. For the box to move without toppling first

   Sliding must occur before toppling, so the sliding force should be less than or equal to the toppling force: $F_{\text{slide}} \le F_{\text{topple}}$

   Substituting, $\mu mg \le \frac{mg\,(a/2)}{(a/2+b)}$

   Cancel $mg$: $\mu \le \frac{a/2}{a/2+b}$

5. Substitute $\mu = 0.4$

   $0.4 \le \frac{a/2}{a/2+b}$

   Rearranging: $0.4\left(\frac{a}{2}+b\right) \le \frac{a}{2}$

   $0.2a + 0.4b \le 0.5a$

   $0.4b \le 0.3a$

   $b \le 0.75a$

6. But the point of application must lie on the box

   Since the cube has side $a$, the highest possible point on the box is its top face, which is at height $a$ above the floor.

   The centre of mass is at height $a/2$, so the maximum physically possible value of $b$ is $b_{\max}=a-\frac{a}{2}=\frac{a}{2}$

   This is less than $0.75a$, so the no-toppling condition is automatically satisfied for all possible points on the box.

   Therefore, the maximum possible value is $b_{\max}=\frac{a}{2}$

7. Required quantity

   $100\times \frac{b}{a}=100\times \frac{1}{2}=50$

Final Answer

$\boxed{50}$