Step 1:

Let $x$ be the mass of the substance at time $t$ in days

When $t=0,x_0=10mg$ and $\large\frac{dx}{dt}$=0.051mg/day

Step 2:

Now $\large\frac{dx}{dt}\propto x$

It is negative because there is decay

$\therefore\large\frac{dx}{dt}=$$kx$

When $t=0$

$-0.051=k(10)$

$\Rightarrow k=-0.0051$

Step 3:

$\therefore \large\frac{dx}{dt}$$=-0.0051x$

$\large\frac{dx}{x}$$=0.0051dt$

Step 4:

Integrating on both sides we get,

$\int \large\frac{dx}{x}$$=0.0051t+\log c$

$\log x=-0.0051t+\log c$

$\log \large\frac{x}{c}$$=-0.0051t$

$x=ce^{\large -0.0051t}$

Step 5:

When $t=0,x_0=10=c$

$\therefore x=10e^{\large -0.0051t}$

Step 6:

When $x=5mg$,to find $t$

$5=10e^{\large 0.0051t}$

$\therefore \large\frac{1}{2}=e^{0.0051t}$

Or $\log _e\large\frac{1}{2}$$=-0.0051t$

$-\log _e2=-0.0051t$

$t=\large\frac{\log _e2}{0.0051}$

$\log _e2=0.6931$

$\therefore t=\large\frac{0.6931}{0.0051}$

$t=135.9$

Therefore the mass is reduced from 10mg to 5mg in 136 days approximately.