The value of

$(0.16)^{\log _{2.5}\left(\frac{1}{3}+\frac{1}{3^{2}}+\frac{1}{3^{3}}+\ldots . \infty\right)}$

$=0.16$ $\log _{2.5}\left(\frac{\frac{1}{3}}{1-\frac{1}{3}}\right)$$\left[\because S_{\infty}=\frac{a}{1-r}\right]$

$=0.16^{\log _{2}}$ $16^{\log _{2.5}\left(\frac{1}{2}\right)}$

$=(2.5)^{-2 \log _{2.5}\left(\frac{1}{2}\right)}=\left(\frac{1}{2}\right)^{-2}=4$