Givens and Knowns:

\(\displaystyle\rho_{{{a}}}={1.16}{k}\frac{{g}}{{m}^{{{3}}}}\)

\(\displaystyle{D}={12}{m}\)

\(\displaystyle{m}_{{{p}}}={85}{k}{g}\)

Solution:

\(\displaystyle{a}={\frac{{{F}_{{\ne{t}}}}}{{{m}_{{\to{t}}}}}}\)

\(\displaystyle{F}_{{\ne{t}}}={F}_{{{B}}}-{W}\)

Where \(\displaystyle{F}_{{{B}}}\) is the bouyancy force and W is the total weight.

\(\displaystyle{F}_{{{B}}}=\rho_{{{a}}}\times{V}{o}{l}\times{g}={1.16}\times{\frac{{\pi\times{4}\times{6}^{{{3}}}}}{{{3}}}}\times{9.81}={10296.02}\)

\(\displaystyle{m}_{{\to{t}}}={2}\times{m}_{{{p}}}+\rho_{{{h}{e}}}\times{V}{o}{l}={170}+{\frac{{{1.16}}}{{{7}}}}\times{904.78}={319.934}\)

\(\displaystyle{W}={g}\times{m}_{{\to{t}}}={9.81}\times{319.934}={3138.56}\)

(Note: \(\displaystyle\rho_{{{h}{e}}}={\frac{{\rho_{{{a}}}}}{{{7}}}}\))

Now, we can calculate the net force exerted as follows,

\(\displaystyle{F}_{{\ne{t}}}={10296.02}-{3138.56}={7157.46}\)

\(\therefore a=\frac{7157.46}{319.934}=22.372\)