You have correctly found the common difference.
The $k$th term of an arithmetic sequence with initial term $a_1$ and common difference $d$ is
$$a_k = a_1 + (k - 1)d$$
Since we are given that $a_3 = 10$ and $a_5 = 16$, we obtain
\begin{align*}
a_3 & = a_1 + (3 - 1)d = 10\\
a_5 & = a_1 + (5 - 1)d = 16
\end{align*}
which is a system of two linear equations in two variables. Subtracting the first equation from the second yields $2d = 6 \implies d = 3$, as you found. Substituting $d = 3$ in the equation for $a_3$ yields $a_1 = 4$, as you found in the comments.
The $n$th partial sum (the sum of the first $n$ terms) of an arithmetic series
\begin{align*}
S_n & = \sum_{k = 1}^{n} a_k\\
& = \sum_{k = 1}^{n} [a_1 + (k - 1)d]\\
& = \sum_{k = 1}^{n} a_1 + \sum_{k = 1}^{n} (k - 1)d\\
& = a_1\sum_{k = 1}^{n} 1 + d\sum_{k = 1}^{n} (k - 1)\\
& = na_1 + \frac{d(n - 1)n}{2}
\end{align*}
from which you can find $S_{20}$ by substituting $20$ for $n$, $3$ for $d$, and $4$ for $a_1$.
Alternatively,
\begin{align*}
S_n & = \sum_{k = 1}^{n} a_k\\
& = a_1 + a_2 + a_3 + \cdots + a_{n - 2} + a_{n - 1} + a_n\\
& = a_1 + [a_1 + d] + [a_1 + 2d] + \cdots + [a_1 + (n - 3)d] + [a_1 + (n - 2)d] + [a_1 + (n - 1)d]
\end{align*}
Since we obtain the same sum if we write the terms in reverse order, we obtain
\begin{align*}
S_n & = a_n + a_{n - 1} + a_{n - 2} + \cdots + a_3 + a_2 + a_1\\
& = [a_1 + (n - 1)d] + [a_1 + (n - 2)d] + [a_1 + (n - 3)d] + \cdots + [a_1 + 2d] + [a_1 + d] + a_1
\end{align*}
Adding the two expressions for $S_n$ yields
\begin{alignat*}{10}
S_n & = & a_1 & + & [a_1 + d] & + & \cdots & + & [a_1 + (n - 1)d] & + & [a_1 + (n - 1)d]\\
S_n & = & [a_1 + (n - 1)d] & + & [a_1 + (n - 2)d] & + & \cdots & + & [a_1 + d] & + & a_1\\ \hline
2S_n & = & [2a_1 + (n - 1)d] & + & [2a_1 + (n - 1)d] & + & \cdots & + & [2a_1 + (n - 1)d] & + & [2a_1 + (n - 1)d]
\end{alignat*}
Since there are $n$ columns in that sum, we obtain
\begin{align*}
2S_n & = n[2a_1 + (n - 1)d]\\
2S_n & = n[a_1 + a_1 + (n - 1)d]\\
2S_n & = n(a_1 + a_n)\\
S_n & = \frac{n(a_1 + a_n)}{2}
\end{align*}
from which you can find $S_{20}$ by substituting $4$ for $a_1$ and using the formula $a_n = a_1 + (n - 1)d$ with $n = 20$ and $d = 3$ to find $a_{20}$.
