A tube light of $60V, 60W$ rating is connected across an AC source of $100V$ and $50Hz$ frequency. Then,

$P_R=V_{R}i$

$\therefore i=\frac{P_R}{V_R}=\frac{60}{60}=1A$

Now, $V_L=\sqrt{V^2-V_R^2}=\sqrt{{\left(100\right)}^2-{\left(60\right)}^2}=80V=i{X}_L=i\left(2\mathrm{\pi fL}\right)$

Then, $L=\frac{80}{2\mathrm{\pi fi}}=\frac{80}{\left(2\mathrm{\pi }\right)\left(50\right)\left(1\right)}=\frac{4}{5\mathrm{\pi }}A$

If we connect another resistance R in series, then voltage drop across it should be 40V, so that remaining 60V is used by the tube light.

$R=\frac{V}{i}=\frac{40}{1}=40\Omega$.