In a system called star system, $ 1starKg = {10^{23}}Kg $ , $ 1starmeter = {10^8}m $ , $ 1starsecond = {10^{23}}\sec $ , then calculate the value of $ 1Joule $ in this system.

Hint: In order to solve this question, we are going to first express the SI units into star unit form, then, writing the term $ 1Joule $ in terms of the units $ Kg{m^2}{\sec ^{ - 2}} $ , we can finally write it in terms of the units $ starKgstarmete{r^2}star{\sec ^{ - 2}} $ by using the conversions obtained above.
Formula used: The work done is obtained by the formula
 $ W = m \times a \times s $
Where $ m $ is the mass, $ a $ is the acceleration and $ s $ is the distance
In terms of the units
 $ J = Kg \times meter{\sec ^{ - 2}} \times meter \\
   \Rightarrow J = Kgmete{r^2}{\sec ^{ - 2}} \\ $

Complete step by step solution:
 Let us start solving the question by writing the conversion of the system of the units as given in the question,
 $ 1starKg = {10^{23}}Kg \\
  1starmeter = {10^8}m \\
  1star\sec = {10^{23}}\sec \\ $
If we calculate the inverted units in this system by dividing both the sides of the equations by the factors same as given at the right hands sides
 $ 1Kg = {10^{ - 23}}starKg \\
  1m = {10^{ - 8}}starmeter \\
  1\sec = {10^{ - 23}}star\sec \\ $
As we know that the joule in SI system can be expanded by the equation
 $ 1Joule = 1Kg{m^2}{\sec ^{ - 2}} $
Thus, in that system of units, we get joule by using the conversions of $ meter $ to $ starmeter $ , $ Kg $ to $ starKg $ , $ m $ to $ starmeter $
 $ 1Joule = {10^{ - 23}}starKg \times {10^{ - 16}}starmete{r^2} \times {\left( {{{10}^{23}}} \right)^2}star{\sec ^{ - 2}} \\
   \Rightarrow 1Joule = {10^7}starKgstarmete{r^2}star{\sec ^{ - 2}} $

Note:
It is important to note that if conversion of one unit to the other is given, then the opposite can be obtained by simply dividing the coefficient of one side by both of the sides. For example: $ 1m = 100cm $ gives the conversion of the unit meter to centimeter
Then the conversion centimeter to meter can be obtained by dividing both sides by $ 100 $
 $ 1cm = \dfrac{1}{{100}}m $ .