Under ideal growth conditions, E.coli strain BL21(DE3) can express up to 30% of its total protein content from an inducible T7 expression vector....

Since the E.coli cells can express only a maximum of 30% of its total protein content, we first have to figure out the total protein content in the given concentration of cells.

It is given that 5 x 10^9 E.coli cells have a total protein yield of 45 `mug`. The given concentration of the E.coli culture is 4.2 x 10^9 cells/ml. In other words, 1 ml of culture contains 4.2 x 10^9 cells. 

Using the given data on total protein yield,

5 x 10^9 cells contain 45 `mug` ,

4.2 x 10^9 cells will contain a total protein content of: 

`45/(5 xx 10^9) xx 4.2 xx 10^9` =  37.8 `mug`.

So, the given culture contains 37.8 `mug`/ml of protein. However, only 30% of it can be expressed.

Thus, the theoretical protein yield = 30% of 37.8 `mug`/ml = 11.34 `mug`/ml

=  1.1e1 `mug`/ml  (in scientific notation).

Hope this helps.