Step 1: Think about the water column drained.
As the water table drops by 6 m over 100 km$^2$, the aquifer effectively drains a slab of rock 6 m thick across that whole area. The volume of that slab is
\[ V_{slab}=A\times\Delta h \]
Step 2: Work out the slab volume.
$1$ km$^2$ is $10^6$ m$^2$, so $A=100\times10^6=10^8$ m$^2$. With $\Delta h=6$ m,
\[ V_{slab}=10^8\times6=6\times10^8\ \text{m}^3 \]
Step 3: Compare pumped water to the slab volume.
Only part of that slab's pore space actually gave up water; the rest is specific retention. The fraction that gave up water is the specific yield, found by dividing the actual water pumped by the slab volume:
\[ S_y=\frac{V_{pumped}}{V_{slab}}=\frac{2\times10^{7}}{6\times10^{8}} \]
\[ S_y=0.0333=3.33\% \]
Step 4: Add back the water that stayed behind.
Total porosity is the sum of what drained out and what clung to the grains, that is specific yield plus specific retention:
\[ n=S_y+S_r=3.33\%+15\%=18.33\% \]
Step 5: Round to the asked precision.
The question wants the answer to one decimal place.
\[ \boxed{18.3\%} \]