Given,

$2x=3y$

$\Rightarrow \frac{x}{3}=\frac{y}{2}$

Now, let $\frac{x}{3}=\frac{y}{2}=k$

$\Rightarrow x=3k \& y=2k$

Now, given set $A=\left\{1, 2, 3, .......,100\right\}$

So, for $k\le 33$ we get,

$\Rightarrow R=\left\{\left(3,2\right),\left(6,4\right),\left(9,6\right),\left(12,8\right),....\left(99,66\right)\right\}$

$\Rightarrow n\left(R\right)=33$

Now, $R_1$ is symmetric relation on $A$ and $R\subset R_1$ so, we need elements $\left\{\left(3,2\right),\left(6,4\right),\left(9,6\right),\left(12,8\right),....\left(99,66\right), \left(2,3\right),\left(4,6\right),\left(6,9\right), \left(8,12\right),....\left(66,99\right)\right\}$ in $R_1$

Hence, the minimum number of elements in $R_1$ is $33+33=66$