$\because$ The zero of $x-\frac{1}{2}$ is $\frac{1}{2}$                    $\left[\because x -\frac{1}{2}=0 \Rightarrow x =\frac{1}{2}\right]$

and  $p ( x )= x ^{3}+3 x ^{2}+3 x +1$

$\therefore $  For divisor $= x -\frac{1}{2},$ remainder is given as

$p \left(\frac{1}{2}\right)=\left(\frac{1}{2}\right)^{3}+3\left(\frac{1}{2}\right)^{2}+3\left(\frac{1}{2}\right)+1=\frac{1}{8}+\frac{3}{4}+\frac{3}{2}+1$

$=\frac{1+6+12+8}{8}=\frac{27}{8}$

Thus, the required remainder $=\frac{27}{8}$.