The Insane Sultan of Slickcrudistan: Calculating Currency Equilibrium
Using a simple textbook-approach, I will provide one possible answer here, showing that in this approach, the one thing that needs to be specified is "velocity of money" as regards sand dollars. 

I will use also the numerical values provided in the question, after cutting out six zeros.
I assume that the Law of One Price holds, so

$$P_s\cdot S_{USD/s} = P_{USD} \tag{1}$$

$P_s$ is "the price level" related to sand dollars, $P_{USD}$ is the price level related to USD, and $S_{USD/s}$ is the exchange rate we wish to find ("how many USD per one sand dollar").

While the price levels are usually price-indexes, here we are looking at a single good, oil, so we can take the $P's$ to mean "price of oil".  

The Sultan's oil is the only output that can be bought with sand dollars. And we do know the value of this output in USD

$$P_{USD} \cdot Q_s = P_{USD} \cdot 20 = 1,000 \;\text{USD} \tag{2} $$

Using the Law of One Price, we get

$$P_{USD} \cdot Q_s = P_s\cdot S_{USD/s}\cdot Q_s \implies S_{USD/s} = \frac {P_{USD} \cdot Q_s}{P_s\cdot  Q_s} \tag{3}$$

(don't simplify).

Assuming the Quantity Theory of Money, we have

$$P_s\cdot Q_s = V_s \cdot \bar M_s \tag{4}$$

where $\bar M_s$ is the fixed amount of sand dollars available, and $V_s$ is the "velocity of money" in relation to sand-dollars (we will return to $V_s$). Inserting $(4)$ into $(3)$ and using also $(2)$ and the fixed value of $\bar M_s$ we have

$$(4),(3),(2) \implies S_{USD/s} = \frac {P_{USD} \cdot Q_s}{V_s\bar M_s} = \frac {1,000}{V_s\cdot 100} = \frac {10}{V_s}$$

So, if we can find a value for $V_s$, we have found the exchange rate we are after. The velocity of money can be thought as "how many times that same unit of currency, say the same paper bill, will be used in transactions during a production cycle".  

The M1 velocity of money for USD is approximately  $V_{USD} \approx 6$, using this value for the sand-dollars also, we get

$$S_{USD/s} = \frac {10}{6} \approx 1.67$$