For the infinite geometric series, $\sum_{n=0}^{\infty} ar^n=\frac{a}{1-r}=57 \qquad (|r|<1)$ so $a=57(1-r).$

Also, $\sum_{n=0}^{\infty} a^3r^{3n}=a^3\sum_{n=0}^{\infty}(r^3)^n=\frac{a^3}{1-r^3}=9747.$

Substitute $a=57(1-r)$: $\frac{[57(1-r)]^3}{1-r^3}=9747.$ Since $1-r^3=(1-r)(1+r+r^2),$ we get $\frac{57^3(1-r)^3}{(1-r)(1+r+r^2)}=9747,$ hence $\frac{57^3(1-r)^2}{1+r+r^2}=9747.$

Now compute: $57^3=185193,$ so $\frac{185193(1-r)^2}{1+r+r^2}=9747.$ Divide both sides by $9747$: $\frac{(1-r)^2}{1+r+r^2}=\frac{9747}{185193}=\frac{1}{19}.$ Thus, $19(1-r)^2=1+r+r^2.$

Expand: $19(1-2r+r^2)=1+r+r^2$ $19-38r+19r^2=1+r+r^2$ $18r^2-39r+18=0.$ Divide by $3$: $6r^2-13r+6=0.$ Factor: $6r^2-13r+6=(3r-2)(2r-3)=0.$ So, $r=\frac{2}{3}\quad \text{or} \quad r=\frac{3}{2}.$

Since the series is infinite and convergent, $|r|<1$, so $r=\frac{2}{3}.$

Then $a=57\left(1-\frac{2}{3}\right)=57\cdot \frac{1}{3}=19.$

Therefore, $a+18r=19+18\cdot \frac{2}{3}=19+12=31.$

Checking options: the correct option is $\boxed{31}$ which is Option C.