Assumption about retardation inside the target
Since the bullet penetrates the target and is brought to rest by the resisting force of the target, we take this resisting force to be constant. Hence the bullet moves with constant retardation inside the target.

Given information
Let the initial speed of the bullet on entering the target be $u$.

After travelling $s_1 = 4\text{ cm} = 0.04\text{ m},$ it loses one-third of its velocity.

So its speed becomes $v = u - \frac{u}{3} = \frac{2u}{3}.$

Use the kinematic equation
For motion with constant acceleration, $v^2 = u^2 + 2as.$

Applying this for the first $0.04\,$m: $\left(\frac{2u}{3}\right)^2 = u^2 + 2a(0.04).$

$\frac{4u^2}{9} = u^2 + 0.08a.$

$0.08a = \frac{4u^2}{9} - u^2 = -\frac{5u^2}{9}.$

Hence, $a = -\frac{5u^2}{9\times 0.08}.$

Distance travelled after this before coming to rest
Now the bullet has speed $\dfrac{2u}{3}$ and comes to rest under the same retardation $a$.

Let the additional distance be $s_2$.

Again use $v^2 = u^2 + 2as,$ with final speed $0$, initial speed $\dfrac{2u}{3}$: $0 = \left(\frac{2u}{3}\right)^2 + 2as_2.$

$0 = \frac{4u^2}{9} + 2as_2.$

Substitute $a = -\dfrac{5u^2}{9\times 0.08}$: $0 = \frac{4u^2}{9} - 2\cdot \frac{5u^2}{9\times 0.08}s_2.$

Cancel $\dfrac{u^2}{9}$: $0 = 4 - \frac{10}{0.08}s_2.$

$\frac{10}{0.08}s_2 = 4.$

$s_2 = \frac{4\times 0.08}{10} = 0.032\text{ m}.$

Find $D$
We are given $s_2 = D\times 10^{-3}\text{ m}.$

Since $0.032\text{ m} = 32\times 10^{-3}\text{ m},$ therefore $D = 32.$

Option check
The correct option is: $\boxed{\text{B: }32}$