There's not enough information given to determine. You are dealing with the subject of "Arc Flash" which even many EE's don't understand fully. The main factors are current, voltage, frequency, short-circuit current available, circuit power factor, type of electrodes, duration of arc, fault clearing time, (to name just a few).

Energy level can range from LOW "Barely enough to detect", such as the spark energy you create by walking across a shag carpet, to HIGH "Infinite", such as the energy in a carbon arc searchlight.

Below is a excerpt from an article from "Electrical Construction and Maintenance" (EC&M) magazine that addresses the topic. This is only given for the purpose of showing what's involved in the calculations.

QUOTE:
Both NFPA 70E and IEEE 1584, “Guide for Performing Arc-Flash Hazard Calculations,” prescribe their own methods of calculation for determining the available arc flash energy at a particular piece of equipment. However, for both methods, typical working distances are given in Table 3 of IEEE 1584-2002.

Calculating arc flash energy
There are two distinct mathematical methods of calculating the available arc flash energy present at a specific piece of equipment — both of which are detailed in Annex D of NFPA 70E.

The first method, commonly referred to as the NFPA 70E equation for an arc flash in a cubic box, is:

EMB = 1038.7 DB-1.4738 × tA [0.0093 F2 0.3453 F + 5.9673]

Where EMB is the arc flash energy, DB is the working distance (from Table 3 of IEEE 1584), tA is the duration of the arc, and F is the short-circuit (or fault) current. This equation uses inches for distance measurements, and gives results directly in calories per centimeter squared (cal/cm2).
END QUOTE

Note: You will need to be able to calculate the Unit Of Measure you want-"Joules" - from the dimension "cal/cm2".

ALLAN MAGEE, PE