Let us assume that the motion of the bullet is of constant retardation inside the water.

Initial kinetic energy $\frac{1}{2}m{v}_0^2=90 \mathrm{J}$

$\Rightarrow v_0=30 \mathrm{m} {\mathrm{s}}^{-1}$

Let the retardation is $a$, then after $1 \mathrm{s}$

$\frac{1}{2}m{\left(30-a\times 1\right)}^2=40 \mathrm{J}$

$\Rightarrow 30-a=20$

$\Rightarrow a=10 \mathrm{m} {\mathrm{s}}^{-2}$

Stopping distance for the constant retardation is given by,

$l=\frac{u^2}{2a}=\frac{{30}^2}{20}=45 \mathrm{m}$.

Therefore, the minimum length of the pool required is $45 \mathrm{m}$.