As the current lags the voltage, the circuit must contain a combination of resistance and inductance.

Power factor $\cos \varphi =0.6$

$\Rightarrow \frac{\mathrm{R}}{\sqrt{{\mathrm{R}}^2+{\mathrm{X}}_{\mathrm{L}}^2}}=\text{0.6}$

$\Rightarrow {\mathrm{R}}^2+{\mathrm{X}}_{\mathrm{L}}^2={\left(\frac{\mathrm{R}}{\text{0.6}}\right)}^2$

$\Rightarrow {\mathrm{X}}_{\mathrm{L}}^2=\frac{{\mathrm{R}}^2}{{\left(\text{0.6}\right)}^2}-{\mathrm{R}}^2$

$\Rightarrow {\mathrm{X}}_{\mathrm{L}}^2=\frac{{\mathrm{R}}^2\times \text{0.64}}{\text{0.36}}$

$\therefore {\mathrm{X}}_{\mathrm{L}}=\frac{0.8\mathrm{ }\mathrm{R}}{0.6}=\frac{4\mathrm{R}}{3}$

Now we want the power factor $\cos \varphi =1$

This is the condition of resonance in which

${\mathrm{X}}_{\mathrm{L}}={\mathrm{X}}_{\mathrm{C}}$

$\therefore {\mathrm{X}}_{\mathrm{C}}=\frac{4\mathrm{R}}{3}$

$\Rightarrow \frac{1}{\mathrm{\omega C}}=\frac{4\mathrm{R}}{3}$

$\Rightarrow \mathrm{C}=\frac{3}{4\times 50\times 100}=150 \mathrm{\mu F}$