Answer:
a)
And since the power can't be negative then the solution would be w = 33.34 watts.
b) 


So then the range of voltage would be between 33.22 W and 33.45 W.
c)For this case
since that's the tolerance 1C
d) 

So then we select the smalles value on this case 
e) For this case if we assume a tolerance of
for the temperature and a tolerance for the power input
we see that:

Where a = 33.34 W, f(x) represent the temperature, x represent the input power and L = 203C
Step-by-step explanation:
For this case we have the following function

Where T represent the temperature in Celsius and w the power input in watts.
Part a
For this case we need to find the value of w that makes the temperature 203C, so we can set the following equation:

And we can rewrite the expression like this:

And we can solve this using the quadratic formula given by:

Where a =0.1, b =2.156 and c=-183. If we replace we got:
And since the power can't be negative then the solution would be w = 33.34 watts.
Part b
For this case we can find the values of w for the temperatures 203-1= 202C and 203+1 = 204 C. And we got this:




So then the range of voltage would be between 33.22 W and 33.45 W.
Part c
For this case
since that's the tolerance 1C
Part d
For this case we can do this:


So then we select the smallest value on this case 
Part e
For this case if we assume a tolerance of
for the temperature and a tolerance for the power input
we see that:

Where a = 33.34 W, f(x) represent the temperature, x represent the input power and L = 203C