Answer
A total of $1.845\times 10^{19}\text{ grains}$ are needed.
Work Step by Step
The sum $S_{n}$ of the first $n$ terms can be computed as:
$S_{n}=\displaystyle \sum_{k=1}^{n}a_{1}r^{k-1}=a_{1}\cdot\frac{1-r^{n}}{1-r}$
Since, $a_{1}=1\\ a_{2}=1\cdot 2\\ a_{3}=1\cdot 2^{2}\\ a_{4}=1\cdot 2^{3}....$. This shows a geometric sequence with $a_{1}=1; r=2; n=64$
Now, the sum $S_{n}$ of the first $64$ terms can be computed as:
$$S_{64}=1\cdot \displaystyle \frac{1-2^{64}}{1-2}=\frac{1-2^{64}}{-1}=2^{64}-1\approx 1.845\times 10^{19}\ \text{ grains}$$
