As the resistance and coil or inductor are in series with power supply, therefore the voltage in the inductor will lead the voltage in resistor by $90°,$the phasor diagram is, 

So, $V_{\mathrm{net}}=20 \mathrm{V}=\sqrt{V_{\mathrm{R}}^2+V_{\mathrm{I}}^2}$and given that $V_{\mathrm{R}}=12 \mathrm{V}$

Therefore, $V_{\mathrm{I}}=\sqrt{{20}^2-{12}^2}=16 \mathrm{V}$

Hence, voltage across the coil is, $16 \mathrm{V}$