Consider the circuit shown in the figure. The current I is equal to

Suppose current through different paths of the circuit is as follows.

After applying KVL for loop (1) and loop (2), We get

$28i_1=-6-8\Rightarrow i_1=-\frac{1}{2}A$

$\mathrm{and}54{\mathrm{i}}_2=-6-12\Rightarrow {\mathrm{i}}_2=-\frac{1}{3}\mathrm{A}$

Hence ${\mathrm{i}}_3={\mathrm{i}}_1+{\mathrm{i}}_2=-\frac{5}{6}\mathrm{A}$

As per the question, $\mathrm{I}={\mathrm{i}}_3=-\frac{5}{6}\mathrm{A}$