Final amount of Hg present after $\beta$ decay
$$\ce{^198Au -&gt; ^198Hg + e^{-}  ~~~~~~~~ \beta~ decay}$$

260 hours corresponds to 4 half-lifes (4 x 65 = 260).  After each half-life, one-half of the $\ce{^198Au}$ will have been transformed into $\ce{^198Hg}$.  After 4 half-lifes ((1/2)^4 = 1/16) 1/16 of the gold will remain and 15/16 of the gold will have been transformed into mercury. Therefor, after 260 hours 15/16 gram atom of mercury will be present.  15/16 gram atom of mercury corresponds to (15/16 x 198 = 185.63 gm) 185.63 gm of mercury.

Using your equation we (of course) get the same result. 
$$\ce{N_t=N_o * e^{-kt}~~(I)}$$ where k is the rate constant and t the elapsed time.  For radioactive decay (a first order reaction) 
$$\ce{k=\frac{ln(2)}{t_{1/2}}~~(2)}$$ where $\ce{t_{1/2}}$ is the decay half-life.  Substituting into equation (I) we find
$$\ce{\frac{N_t}{N_o}= e^{-\frac{ln(2)}{t_{1/2}}*4}=1/16}$$