If $$\log_{2}{x}.\log_{\frac{x}{64}}{2}=\log_{\frac{x}{16}}{2}$$. Then x is

$$\log_{2}{x}.\log_{\frac{x}{64}}{2}=\log_{\frac{x}{16}}{2}$$

i.e. $$\frac{log{x}}{log{2}} * \frac{log_{2}}{log{x}-log{64}} = \frac{log{2}}{log{x}-log{16}}$$

i.e. $$\frac{log{x} * (log{x}-log{16})}{log{x}-log{64}}$$ = $$\log{2}$$

let t = log x

Therefore,  $$\frac{t * (t-log{16})}{t-log{64}}$$ = $$\log{2}$$

$$t^2-4*log 2*t = t*log 2-6*(log 2)^2$$

I.e. $$t^2-5*log 2*t-6*(log 2)^2$$ = 0 

I.e. $$t^2-3*log 2*t-2*log 2*t-6*(log 2)^2$$ = 0 

i.e. $$t*(t-3*log 2)-2*log 2*(t-3*log 2)$$ = 0

i.e $$t=2*log 2$$ or $$t=3*log 2$$

i.e $$log x=log 4$$ or $$log x=log 8$$

therefore $$x=4$$ or $$8$$ 

therefore our answer is option 'B'