The given inequality is $\large\frac{x}{3} > \frac{x}{2} $$+1$
Subtracting $\large\frac{x}{2}$ on both sides,
$=> \large\frac{x}{3} -\frac{x}{2} > \frac{ x}{2} $$+1 -\large\frac{x}{2}$
$=> \large\frac{x}{3}-\frac{x}{2} $$ > 1$
$=> \large\frac{2x-3x }{6} $$ > 1$
$ => \large\frac{-x}{6} $$ > 1$
Multiplying both sides of the inequality by a negative number $-6$.
$=> \large\frac{-x}{6} $$ \times -6 < 1 \times -6$
$=> x < -6$
Step 2:
Hence all real number less than $-6$ satisfy the given inequality
The solution set is $(-\infty, -6)$
Hence B is the correct answer.