Net Force $=F1+F2$

Now, let's calculate the forces using the given values:

$F1=(8\mathrm{.}988{10}^9(N)*\frac{m^2}{C^2})*\frac{|1\mathrm{.}650{10}^{-6}c|*|1\mathrm{.}650{10}^{-6}C|}{(0\mathrm{.}06(m){)}^2}$

$=3\mathrm{.}422N$

$F2=(8\mathrm{.}988{10}^9(N)*\frac{m^2}{C^2})*\frac{|1\mathrm{.}650{10}^{-6}c|*|1\mathrm{.}650{10}^{-6}C|}{(0\mathrm{.}22(m){)}^2}$

$=0\mathrm{.}483N$

$F2=(8\mathrm{.}988{10}^9(N)*\frac{m^2}{C^2})*\frac{|1\mathrm{.}650{10}^{-6}C|*|1\mathrm{.}650{10}^{-6}C|}{(0\mathrm{.}22(m){)}^2}$

$=0\mathrm{.}483N$

Net Force $=$ F$1+$ F$2$

$=3\mathrm{.}422N+0\mathrm{.}483N$

$=3\mathrm{.}905N$

Therefore, the force on the charge located at $16\mathrm{.}00cm$ is $3\mathrm{.}905$