= Solution
Use the positive-support convention for the <geometric distribution> and put $q=1-p$. Differentiating the convergent <geometric series> $\sum_{k\geq0}q^k=(1-q)^{-1}$ gives
$$
\sum_{k\geq1}kq^{k-1}=\frac1{(1-q)^2},\qquad
\sum_{k\geq2}k(k-1)q^{k-2}=\frac2{(1-q)^3}.
$$
Consequently
$$
\mathbb EY=p\sum_{k\geq1}kq^{k-1}=\frac1p,\qquad
\mathbb E[Y(Y-1)]=pq\sum_{k\geq2}k(k-1)q^{k-2}=\frac{2q}{p^2}.
$$
Subtracting the square of the <expectation> from the <second moment> yields
$$
\boxed{\mathbb EY=\frac1p,\qquad\operatorname{Var}(Y)=\frac{1-p}{p^2}.}
$$
For the die, this is the <coupon collector problem>. When $r$ scores are still missing, let $W_r$ count rolls up to and including the next newly observed score. Every fresh roll has success <probability> $r/6$, so $W_r$ has a <geometric distribution> of parameter $r/6$. These six successive waiting times are <independent>: conditional on the entire past at the start of a stage, its waiting-time distribution depends only on the number $r$, not on the identities of the missing scores. Iterated conditioning therefore factors their joint distribution. The total number of rolls is $N=\sum_{r=1}^6W_r$.
Using <linearity of expectation> gives
$$
\boxed{\mathbb EN=\sum_{r=1}^6\frac6r
=6\left(1+\frac12+\frac13+\frac14+\frac15+\frac16\right)
=\frac{147}{10}=14.7.}
$$
Using <variance additivity for independent random variables>, or the <coupon collector waiting-time variance> formula just derived by this decomposition, gives
$$
\operatorname{Var}(N)=\sum_{r=1}^6\frac{1-r/6}{(r/6)^2}
=\sum_{r=1}^6\left[\left(\frac6r\right)^2-\frac6r\right]
=\frac{5369}{100}-\frac{147}{10}=\frac{3899}{100}.
$$
Hence the requested <standard deviation> is
$$
\boxed{\operatorname{sd}(N)=\frac{\sqrt{3899}}{10}\approx6.2442.}
$$
The hint rounds the sum of squares from $53.69$ to $53.7$; using that rounded value gives $\sqrt{39}\approx6.245$, consistent at the precision of the hint. The exact computation preserves the slightly more precise value above.
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