Answer
The data in the table supports Kepler's Third Law.
The length of Jupiter's period: $T \simeq 11.89$ years
Work Step by Step
From the table below we see that the data support that ${T^2}/{a^3}$ has the same value for each planetary orbit, hence, it supports Kepler's Third Law.
$\begin{array}{*{20}{c}}
{Planet}&{Mercury}&{Venus}&{Earth}&{Mars}\\
{a\left( {{{10}^{10}}m} \right)}&{5.79}&{10.8}&{15.0}&{22.8}\\
{T\left( {years} \right)}&{0.241}&{0.615}&{1.00}&{1.88}\\
{{a^3}}&{194.11}&{1259.71}&{3375.0}&{11852.4}\\
{{T^2}}&{0.058}&{0.378}&{1.00}&{3.53}\\
{{T^2}/{a^3}}&{0.000299}&{0.0003}&{0.000296}&{0.000298}
\end{array}$
From the table we take $\frac{{{T^2}}}{{{a^3}}} \simeq 0.0003$. Assuming that $a = 77.8 \times {10^{10}}$ m for Jupiter, the length of its period is
$T \simeq \sqrt {0.0003 \times {{\left( {77.8} \right)}^3}} \simeq 11.89$ years
