If all $20$ new hires were women, then we could set up this system of equations, where $n$ is the number of women in 1990 and $N$ is the total number of employees in 1990:
$$\left\{
\begin{array}{rcl}
\frac{n}{N}&=&0.337\\
\frac{n+20}{N+20}&=&0.362
\end{array}
\right.$$
which is the same as the system:
$$\left\{
\begin{array}{rcl}
n&=&0.337N\\
n+20&=&0.362N+7.24
\end{array}
\right.$$
which has solution $N=510.4$, $n\approx172$.
The percentage of women did go up, so at the other extreme, maybe $\lceil0.337\cdot20\rceil=8$ of the new hires were women. In that case, we have the system
$$\left\{
\begin{array}{rcl}
\frac{n}{N}&=&0.337\\
\frac{n+8}{N+20}&=&0.362
\end{array}
\right.$$
which is the same as the system:
$$\left\{
\begin{array}{rcl}
n&=&0.337N\\
n+8&=&0.362N+7.24
\end{array}
\right.$$
which has solution $N=30.4$, $n\approx10.24$.
As you can see, there are many possibilities depending upon how many of those $20$ are women. Seemingly, there are $13$ possibilities stemming from the $13$ choices (8 through 20) that we have for the number of newly hired women. If you have time, and preferably a spreadsheet program, you could investigate how many of these options are not really compatible with the given decimal precision. For example, if $8$ new hires were women then $N$ is somewhere in the neighborhood of $30$ and $n$ is somewhere in the neighborhood of $10$. But no combination of numerators in $\left\{9, 10, 11, 12\right\}$ and denominators in $\left\{29, 30, 31, 32\right\}$ yields a decimal that rounds to $0.337$.
But realistically, my guess is that the instructor accidentally left out some critical piece of information.