Bayes probability excercise

Observe that

$$\mathbb{P}[A \cap B^c]=\mathbb{P}[A]-\mathbb{P}[A\cap B]=0.98-0.15\times 0.96$$

Thus simply, your result is

$$\mathbb{P}[A | B^c]= \frac{0.98-0.15\times 0.96}{0.85}$$

An intuitive way to think about these types of problems. Think of a very large number of pregnant mothers, normalized to $100$ for simplicity. $% P(A)=0.98$ tells us that $98$ of the babies will survive and $2$ will die. $% P(B)=0.15$ tells us that $15$ will have a c-section. $P(A|B)=0.96$ tells us that $96$ percent of the c-section babies will survive, i.e. $0.96\ast 15=14.4$, and thus that $0.6$ of the c-section babies will die. In total, $2$ babies die, so $2-0.6=1.4$ of these must be among the non c-section babies. The survival among non c-section is thus $(85-1.4)/85= 0.983\,53$