$Delta E=-8.8times 10^5$ J

$W=5.1times 10^{5}$ J
so the heat is

$Q=Delta E+W=(5.1times 10^5 J)+(-8.8times 10^5 J)=-3.7times 10^5 J$.

Since he is the one who does the work its sign will be positive.

Now we write:

$$
begin{align*}
W = 4.3 cdot 10^{5}text{ J}
end{align*}
$$

We know that the swimmer gives off $1.7 cdot 10^{5}text{ J}$ of heat.

Since he releases (loses) the heat the sign will be negative.

So we write:

$$
begin{align*}
Q = – 1.7 cdot 10^{5}text{ J}
end{align*}
$$

Now for the change in thermal energy we write the First Law of Thermodynamics:

$$
begin{equation*}
Delta E = Q – W
end{equation*}
$$

Plugging in the numbers:

$$
begin{align*}
Delta E &= – 1.7 cdot 10^{5}text{ J} – (4.3 cdot 10^{5}text{ J}) = -6 cdot 10^{5}text{ J} \
Delta E &= -6 cdot 10^{5}text{ J}
end{align*}
$$

$$
Delta E=Q-W=(50 J)-(50 J)=0 J
$$