Using, I = K tan $\theta$, we get $\frac{12}{(4 + 2)} $ K = tan 60$\circ$
or 2 = K $\times$ 1.7321 or K = $\frac{2}{1.7321}$
Again, for new situation $\frac{12}{(2 + R)} $ = K tan 30$^\circ$ = $\frac{2}{1.7321} \times 0.5774$ = 0.67
or $\frac{12}{0.67}$ = 2 + R
or R = $\frac{12}{0.67}$ -2 or R = 18 - 2 = 16 $\Omega$