This is false: there exist extensions $0 \to \mathcal K(\ell^2(\mathbb N)) \to A \to B \to 0$ with $B$ being (separable and) exact and $A$ non-exact. But the results underpinning this are deep. Examples can be found in the book of Brown and Ozawa, Theorem 13.4.1 (see Remark 13.4.2 for why this provides counter examples).

Here is a slightly different and more detailed explanation: By a theorem of Kirchberg, every exact $C^\ast$-algebra is locally reflexive, and by a theorem of Effros-Haagerup, if $A$ is separable, locally reflexive and $I$ is a nuclear two-sided closed ideal in $A$, then there exists a completely positive splitting $A/I \to A$. So in an extension $0\to I \to A \to B \to 0$ for which $I$ is nuclear and $B$ is separable, exact, then $A$ is exact (if and) only if the extension has a completely positive splitting. 

In the case where $I = \mathcal K(\ell^2(\mathbb N))$, (and $A, B$ are unital), it follows from basic Brown-Douglas-Filmore theory that every unital extension $0 \to \mathcal K(\ell^2(\mathbb N)) \to A \to B \to 0$ has a completely positive splitting if and only if $\mathrm{Ext}(B)$ is a group. So any example where $B$ is exact and $\mathrm{Ext}(B)$ is not a group, provides a counterexample. Kirchberg showed (which essentially boils down to Theorem 13.4.1 that I mentioned above) that if $B$ is separable, exact, non-nuclear and QWEP, and $C$ is the unitisation of $C_0((0,1], B)$, then $\mathrm{Ext}(C)$ is not a group, and therefore there exists a unital extension $0 \to \mathcal K(\ell^2(\mathbb N)) \to A \to C \to 0$ where $A$ is not exact even though $C$ is. So you can take any separable, exact, non-nuclear, QWEP $C^\ast$-algebra $B$ and construct a counterexample, e.g. $B= C^\ast_r(\mathbb F_2)$.

A different (also very deep) counterexample comes from Haagerup-Thorbjørnsen who show that $\mathrm{Ext}(C^\ast_r(\mathbb F_2))$ is not a group. So there exists a unital extension $0\to \mathcal K(\ell^2(\mathbb N)) \to A \to C^\ast_r(\mathbb F_2) \to 0$ where $A$ is not exact.