Answer
$\dfrac{189}{4}$
Work Step by Step
The sum of the first $n$ terms can be computed as:
$S_n=\dfrac{a_1(1-r^{n})}{1-r}...(1)$
where, $a_1$ is first term and $r$ is the common ratio, $r$ and can be computed as the quotient of a term and the term preceeding it. In a geometric series the ratio between consecutive terms is constant.
We can see from the given series that $a_1=48 \left(\dfrac{1}{2}\right)^{1}=24$ and $r=\dfrac{1}{2}$
Plug $6$ for $n$ and $24$ for $a_1$ and $\frac{1}{2}$ for $r$ in the equation $(1)$ above to obtain:
$S_6=\dfrac{24\left[1-\left(\dfrac{1}{2}\right)^{6}\right]}{1-\dfrac{1}{2}}=\dfrac{24\left[1-\left(\dfrac{1}{64}\right)\right]}{1-\dfrac{1}{2}}=\dfrac{189}{4}$
Thus, the sum of the first six terms is: $S_6=\dfrac{189}{4}$.
