Step 1:

Let $X$ be the random variable denoting the number of defective fuses in a box of 200 fuses.

Probability of a defective fuse P=0.02

$\lambda=np$

$\quad=200\times 0.02$

$\quad=4$

$\therefore X\sim P(4) $

$\Rightarrow P(X=x)=\large\frac{e^{-4}4^4}{4!}$$\qquad x=0,1,2......$

Step 2:

Probability that exactly 4 fuses are defective

$P(X=4)=\large\frac{e^{\Large -4}{4^4}}{4!}$

$\qquad\qquad=0.0183\times \large\frac{32}{3}$

$\qquad\qquad=0.0061\times 32$

$\qquad\qquad=0.1952$