Work in terms of the time constant $\tau=L/R$. The current (and hence the coil's magnetic field) reaches a fraction $f=0.80$ of maximum when $f=1-e^{-t/\tau}$, so $e^{-t/\tau}=1-f=0.20$.
Inverting, $t/\tau=\ln\!\left(\dfrac{1}{0.20}\right)=\ln 5=1.609$, which means $\tau=\dfrac{t}{1.609}=\dfrac{20\,\mu\text{s}}{1.609}=12.43\,\mu\text{s}$.
Since $\tau=L/R$, the resistance is $R=\dfrac{L}{\tau}=\dfrac{2.0\times10^{-3}}{12.43\times10^{-6}}=160.9\,\Omega$. Rounding gives $R\approx160\,\Omega$, matching option (D). As a sanity check, a smaller resistance would give a longer $\tau$ and slower rise, so 160 ohm is consistent with reaching 80 percent in only $20\,\mu\text{s}$.
\[\boxed{R\approx160\,\Omega}\]