The time period of simple pendulum depends onm length of spring (s) and acceleration due to gravity (g)
$$[l]$$ = length = $$[L]$$
$$[g]$$ = acceleration = $$[LT^{-3}]$$
$$[T]$$ = Time period = $$[T]$$
Let $$[T]\propto [L]^a[LT^{-2}]^6$$
comparing LHS and RHS
$$a+b=0,a=-b$$
$$-2b=1,b=-1/2$$
So, $$a=-1/2$$
$$T\propto (8)^{1/2}(g)^{-1/2}$$
$$T\propto \sqrt{\frac{l}{g}}$$
$$T=211\sqrt{\frac{l}{g}}$$
$$2:11$$