A plank of thickness x can just stop a bullet moving with a speed of $\sqrt{3}m{s}^{-1}$. The speed of the bullet that can be just stopped by a plank of thickness 3x is

Let arrange resistance offered by plank be F
$\therefore Fx=\frac{1}{2}m{v}^2\to \left(1\right)$
The speed of the bullet stopped by a plank of thickness 3x is given by $\mathrm{F}\left(3\mathrm{x}\right)=\frac{1}{2}{\mathrm{mv}}_1^2\to \left(2\right)$
From equation (1) and (2) we get
$3\times \frac{m{v}^2}{2}=\frac{1}{2}m{v}_1^2\Rightarrow v_1=3m{s}^{-1}$