Copyright © University of Cambridge. All rights reserved.
The recipe uses a tin of radius 10cm and depth 7.5cm. This has a volume of
$$\pi\times10^2\times7.5 cm^3 = 2400 cm^3$$
We don't know the depth of the 23cm round tin. If its depth is also 7.5cm, then its volume is given by:
$$ \pi\times11.5^2\times7.5 cm^3 = 3100 cm^3$$
which would be fine.
The limit for the depth of this tin could be found by trial and error, or you could rearrange the formula for the volume of a cylindrical tin to find the height which gives a volume of 2400 cm3.
$$ \pi\times11.5^2\times h cm^3 = 2400 cm^3$$ $$h = \frac{2400}{\pi\times11.5^2} cm = 5.8 cm$$
So depending on the depth of the 23cm round tin, all could be well and Toby gets his cake!
The volume of the square tin is $15^2\times6cm^3=1350cm^3$, which isn't large enough. For a large enough square tin of the same depth, we need:
$$l^2\times 6 cm^3 = 2400 cm^3$$
$$l = \sqrt{\frac{2400}{6}} cm = 20 cm$$
where l is the length of the side of the tin.
If you think of the diagonal of the square tin as being equivalent to the diameter of a round tin, the length of the diagonal, d, is:
$$d^2 = 2\times l^2$$
$$d = \sqrt{2\times20^2} = 28.3 cm$$
so quite a bit longer than the diameter of the 20cm round tin!